Question

in a triangle with sides ( a = 6 ), ( b = 8 ), and ( c = 10 ), which formula should be used to find angle ( b )?
( cos b=\frac{a^{2}+c^{2}-b^{2}}{2ac} )
( \frac{b}{cos b}=\frac{c}{cos c} )
( a^{2}+b^{2}=c^{2} )
( \frac{a}{sin a}=\frac{b}{sin b} )

Explanation:

Step1: Recall the Law of Cosines

The Law of Cosines formula is $\cos B=\frac{a^{2}+c^{2}-b^{2}}{2ac}$ for a triangle with sides \(a\), \(b\), \(c\) and angle \(B\) opposite side \(b\).

Step2: Analyze each option

• Option a: $\cos B=\frac{a^{2}+c^{2}-b^{2}}{2ac}$ is the Law of Cosines formula for finding \(\cos B\).
• Option b: \(\frac{b}{\cos B}=\frac{c}{\cos C}\) is not a valid trigonometric formula for triangles.
• Option c: \(a^{2}+b^{2}=c^{2}\) is the Pythagorean theorem (used for right - angled triangles to check the relationship between sides, not for finding angles).
• Option d: \(\frac{a}{\sin A}=\frac{b}{\sin B}\) is the Law of Sines (used when we know the ratio of sides and opposite angles, but here we want a formula specifically for \(\cos B\) in terms of the sides).

Answer:

A. \(\cos B=\frac{a^{2}+c^{2}-b^{2}}{2ac}\)